Counting is modelling, not just arithmetic

A counting principle is a rule for finding the number of possible outcomes in a structured situation without writing out every outcome one at a time. That is the point of this part of the syllabus: show the situation clearly, then let the structure count for you.

The first rule you’ll use most often is the multiplication principle. It says that if one choice can be made in $n$ ways and, after that, a second choice can be made in $m$ ways, then the two-stage process can be completed in $nm$ ways, where $n$ is the number of choices at the first stage and $m$ is the number of choices at the second stage. The same idea works for more than two stages: multiply the number of choices at each stage, as long as each stage has been counted correctly.

A tree diagram is often the clearest way to see this. You don’t draw it because the examiner enjoys trees; you draw it because it makes you ask, “How many branches come next?”

Factorials

A factorial is the product of all positive integers from a given positive integer down to $1$. For a positive integer $n$,

Also, $0! = 1$. That last fact is not a trick; it keeps the counting formulae working neatly at the edges.

For example, arranging all $5$ different students in a row gives

$$5! = 5\cdot 4\cdot 3\cdot 2\cdot 1 = 120$$

arrangements, since there are $5$ choices for the first position, then $4$ for the next, and so on.

Permutations: order matters

A permutation is an ordered selection of objects from a set. The key word is “ordered”: chair, secretary, treasurer is a different assignment from secretary, chair, treasurer, even if the same three people are involved.

The number of permutations of $r$ objects chosen from $n$ distinct objects is

This is also written as $P(n,r)$ on some calculators.

So if $3$ prize positions are awarded from $8$ competitors, the number of possible ordered outcomes is

$${}^8P_3=\frac{8!}{(8-3)!}=8\cdot 7\cdot 6=336$$

You can see the multiplication principle inside the formula: $8$ choices for first place, $7$ for second, $6$ for third.

Combinations: order does not matter

A combination is an unordered selection of objects from a set. If a group of $4$ students is chosen for a photograph, it is the same group no matter which order you say the names.

The number of combinations of $r$ objects chosen from $n$ distinct objects is

This is also written as $C(n,r)$ or $\binom{n}{r}$.

It helps to understand the link between the two formulae rather than memorise them as separate facts:

$${}^nP_r = {}^nC_r\,r!$$

Here $r!$ counts the ways to arrange the $r$ selected objects. In words: choose the group, then arrange the group. That is why permutations are larger than combinations whenever $r>1$.

Choosing the correct model

The classroom test is simple:

• If changing the order changes the outcome, use a permutation or a staged multiplication argument.
• If changing the order does not change the outcome, use a combination.
• If the situation has several independent categories, count each category and multiply.

For instance, choosing $2$ students from a group of $9$ and $3$ students from a different group of $7$ gives

$${}^9C_2\times {}^7C_3$$

because the choices from the two groups are both needed.

For “at least one” conditions, counting the complement is often cleaner. A complement is the set of outcomes not satisfying the condition within the full set of possible outcomes. For example, if a team of $4$ is chosen from $6$ juniors and $5$ seniors, and the team must contain at least one of each, count all teams and subtract the forbidden all-junior and all-senior teams:

$${}^{11}C_4-{}^6C_4-{}^5C_4$$

This is a modelling choice, not a shortcut for its own sake. “At least one” can mean several cases; the complement often turns those several cases into one subtraction.

What is not required here

The syllabus deliberately excludes two common extensions:

• permutations where some objects are identical;
• circular arrangements.

So if you have seen formulae such as dividing by repeated factorials, or fixing one person in a circle, keep those ideas separate. They are not required for this content statement.

Lottery numbers and large counts

Counting principles can turn ordinary choices into very large numbers. In a lottery-style setting, the number of possible tickets can be enormous even when each ticket looks simple. Here the mathematics meets ethics: if a seller advertises a prize without helping buyers understand how many tickets are possible, the numerical truth is technically present but practically hidden.

That is a good example of what “understanding” means in mathematics. Getting the correct value of ${}^nC_r$ is one level; interpreting the size of that number in context is the deeper level.

From finite expansion to infinite expansion

The SL binomial theorem gives a finite expansion when the index is a non-negative integer. Now we use the same idea with fractional and negative indices. The trade-off is simple: the expansion usually becomes infinite, and it only works for certain values of the variable.

The extended binomial theorem is

This expansion is valid when $|x|<1$.

Watch the letter $n$ here. In the ordinary binomial theorem, $n$ is usually a non-negative integer, so the expansion stops. In the extended theorem, $n$ may be fractional or negative, and the expansion normally keeps going.

A rational number is a number that can be written as $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$. So indices such as $\frac{1}{2}$, $-1$, and $-\frac{3}{2}$ are all allowed.

Rewriting the binomial first

Most questions do not arrive neatly in the form $(1+x)^n$. First rewrite the expression so that the bracket begins with $1$:

This rewrite is doing real work. It shows you the “small” quantity that plays the role of $x$. The condition for convergence becomes

$$\left|\frac{b}{a}\right|<1$$

For example,

$$(4+x)^{-1}=4^{-1}\left(1+\frac{x}{4}\right)^{-1}$$

Now the expansion is valid when

$$\left|\frac{x}{4}\right|<1$$

which means $|x|<4$.

Working term by term

For a fractional or negative index, write the coefficient pattern with care. For example,

$$(1+x)^{-2}=1+(-2)x+\frac{(-2)(-3)}{2!}x^2+\frac{(-2)(-3)(-4)}{3!}x^3+\cdots$$

So

$$(1+x)^{-2}=1-2x+3x^2-4x^3+\cdots$$

The signs are not guessed. They come from the factors $n$, $n-1$, $n-2$, and so on. I always tell students to write one more unsimplified line than they think they need. It prevents most sign errors.

For a square-root style example,

$$(1+x)^{\frac{1}{2}}=1+\frac{1}{2}x+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}x^2+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}x^3+\cdots$$

so

$$(1+x)^{\frac{1}{2}}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+\cdots$$

Approximations

The extended binomial theorem is especially useful for approximating numbers. To approximate $\sqrt{2}$, put it in a form where the expression inside the square root is close to $1$. For instance,

$$\sqrt{2}=\sqrt{\frac{9}{4}\cdot\frac{8}{9}}=\frac{3}{2}\left(1-\frac{1}{9}\right)^{\frac{1}{2}}$$

Here $x=-\frac{1}{9}$, which is small enough because $\left|-\frac{1}{9}\right|<1$. Using the first few terms,

$$\left(1-\frac{1}{9}\right)^{\frac{1}{2}}\approx 1-\frac{1}{18}-\frac{1}{648}-\frac{1}{11664}$$

Therefore,

$$\sqrt{2}\approx \frac{3}{2}\left(1-\frac{1}{18}-\frac{1}{648}-\frac{1}{11664}\right)$$

This is not meant to beat your calculator at producing a decimal. The point is that a general algebraic expansion gives a controlled approximation to a number.

Link with power series

A power series is an infinite sum of terms involving increasing powers of a variable. The extended binomial theorem is one of the first power series you meet in the course. Later, in power series expansions, the same idea appears more widely: replace a difficult function by a polynomial-like expression that is valid near a particular value.

Pascal’s triangle and attribution

The coefficients in the ordinary integer-index binomial theorem are often displayed in “Pascal’s triangle”. Historically, those patterns were known in several mathematical cultures before Pascal, including work associated with Yang Hui in China. That matters: mathematics is full of shared discoveries, rediscoveries and different representations of the same structure.

So when we name a result after one person, treat the name as a convenient label, not a complete history. In TOK language, this is a reminder that mathematical knowledge is not only a list of correct results; it is also shaped by communication, preservation and attribution.

What is not required here

You are not required to prove the binomial theorem in this content statement. You do need to know how to use the extended expansion, how to rewrite $(a+b)^n$ into the required form, and how to state the convergence condition.